There are two key words that will help us with this question. One is dwelling unit, and the other is ground-fault circuit interrupter protection for personnel. There are two ways to go about finding this answer. One is to remember that Art. 210 has a ton of information on receptacles used in dwelling units. Knowing that, you can look at your tabs and see a tab labled 210.8 GFCI Protection. That's a good place to turn real quick, and looking at 210.8(A), you'll notice the heading is Dwelling Units, which is the other key word. You'll also notice 210.8(A) has a list of places GFCIs are required, and living room is the only item not in that list. The other way to find this answer, is to go to the index and find one of the key words. Dwelling Units isn't in the index, but Ground-fault circuit interrupters is. Scanning down the list, we see our other keyword Dwelling Units in several locations, and they all point us to 210.8(A) as well. Another thing to keep in mind with GFCIs, is they are typically required in locations that have the potential for contact with water. That way, if you're not able to find the answer, you can make an educated guess on the question, increasing your odds of picking the correct answer.

This one is a little tough, because there are a few things involved. This question requires some familiarity with the code that isn’t as simple as looking up a section and finding the answer. Firstly, it is usually assumed that lights are a continuous load. The question explicitly lets us know it's a continuous load by stating the fixtures operate for 8 hours per day. Looking up the definition of continuous load in Art. 100, we find out it is “A load where the maximum current is expected to continue for 3 hours or more.” This would apply to most lighting applications. Why is this important? Well, let’s look in the index under “Continuous Load”. We’ll find here are two spots in Art. 210 that this references. Art. 210 deals with Branch Circuits and since light fixtures fall under the umbrella of Branch Circuits, that’s a good place to start. Looking at 210.19, we find it deals with conductors and 210.20 deals with Overcurrent Protection. We know this is a 20 amp circuits, which means it’s protected by a 20 amp breaker. Let’s dig into 210.20 a little more.

Ah, 210.20(A) deals with Continuous and Noncontinuous Loads. That’s what we’re looking for. Reading through it, we find the breaker has to be sized at no less than 100% of the noncontinuous load and 125% of the continuous load. For this circuit, there is zero noncontinuous load, so this breaker has to be sized at no less than 125% of the load. What does that mean?!? It means a load of 100A needs a breaker rated at least 125A.

100A X 125% = 125A

But we’re dealing with a breaker that’s 20A. What’s the max continuous load we can put on a 20A breaker? Well, let’s look at the formula.

Continuous Load X 125% = Minimum Breaker/Fuse size

Okay, let’s go back to 8th grade and pull out our algebra skills. We’re looking for the max continuous load? So, if we divide both sides of our equation by 125%, we can cancel out the 125% on the left side, leaving Continuous Load all by it’s lonesome, and get Min Breaker Size / 125%.

Our new equation looks like this: Max Continuous Load = Breaker Size / 125%.

Let’s make that a little prettier and convert our percentage to a decimal. So, our equation is Max Continuous Load = Breaker Size / 1.25.

We’re almost there. Our breaker size is 20A. Let’s use our new Max Continuous Load formula. 20A / 1.25 = 16A. Now we know Max Continuous Load is 16A, and knowing is half the battle.

Now, how many amps does each of these fixtures draw? Well, using our handy-dandy Power Equation, we know that Power(P) = Current(I) X Voltage(E), or P=IE, which I like better, because it reminds me of pie… mmm, pie. So, using the same division property of equality (that’s a fancy way of saying divide something on both sides of an equation), we can rewrite our Power Formula to find current. So, I = P/E.

We know the power draw of each fixture is 100W, and the voltage is 120V, so to find current, we simply use the formula I = 100W / 120V, which tells us each fixture draws .833A. Almost finished now. We simply take our Max Continuous Load of 16A and divide by hour .833A per fixture. 16A / .833A per fixture = 19.2 fixtures. Since we can’t have a fraction of a fixture, and we can’t exceed 19.2 fixtures, because it’s a max value, we have to drop our fixture count to 19 fixtures. Let’s double check ourselves and figure out how many amps 19 fixtures will draw. 19 fixtures X .833A per fixture = 15.827 A, which is below our Max Continuous Load of 16A.

So, the answer is 19 fixtures. And now you know why engineers make the big bucks.

We have three key words here.

1. Branch Circuit
2. Dwelling Units
3. Small-appliance Branch Circuits
4. Kitchen

In general, the keyword Branch Circuit lets us know we're going to be in Art. 210. Let's look in the index for Branch Circuit. Looks like a lot of stuff. Let's scan down and see if any of our other keywords are there. No Dwelling Units, no Kitchen, but we do have Small-appliance. Let's go to 210.11(C)(1).

Reading through it, we see two or more 20-ampere small-appliance branch circuits shall be provided for all receptacle outlets specifed by 210.52(B). It's looking like 2 might be the answer, but let's go to 210.52(B) to see what it says.

Glancing at 210.52(B)(1) we see the word Kitchen. There's our fourth key word. This section mentions the two or more 20-ampere small-appliance branch circuits required by 210.11(C)(1). Again, it seems in a kitchen, the answer is 2

Scanning at the headings, we'll also notice 210.52(B)(3) Kitchen Receptacle Requirements. Our fourth keyword again. Reading through, we see not fewer than two small-appliance branch circuits.

Three confirmation on two small-appliance branch circuits. Our answer is 2.